on generalized choral sequences

8/6/2019 On Generalized Choral Sequences

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On Generalized Choral Sequences

Joel Reyes Noche

Department of Mathematics and Natural SciencesAteneo de Naga UniversityNaga City, Camarines Sur

email: [email protected]

Abstract

Generalized choral sequences are innite binary words ( c n )n 0 dened by c3 i + r 0 =0 , c3 i + r 1 = 1 , and c3 i + r c = c i (where the r s are distinct xed elements of {0, 1, 2})for all non-negative integers i . We present some of their properties. In particular, weshow how each is generated by a deterministic nite automaton with output and howeach is a xed point of a uniform morphism. We also look at their subword complexityand Lyndon factorizations.

1 Introduction

We use denitions and notation from Allouche and Shallit [1]. Denote by N the set of non-negative integers and by Z + the set of positive integers. A word is a concatenation (asequence) of letters chosen from an alphabet (a non-empty set of letters). A nite word hasa nite length (the number of letters it contains) and is denoted by a lowercase italic letter.The length of a nite word w is denoted by |w| . The empty word , denoted by , has a lengthof zero. A one-sided right-innite word (which we will simply call an innite word ) is a mapfrom N to an alphabet and is denoted by a lowercase boldface letter. We call a word overthe alphabet {0,1 } a binary word .

Concatenation of words is denoted by the juxtaposition of their symbols. For example,if w = 0 and x = 01 , then wx = 001 , xw = 010 , w1 = x, and wwxwx = w(wx)2 = 0 3 10 2 1 .We dene w1 = w and w0 = for any nite word w. If x is a nite non-empty word, thenx is the innite word xxx .

A word y is a subword of w if there exist x and z such that w = xyz . If x = , then y isa prex of w. If z = , then y is a suffix of w. If x = and z = , then y is a proper prex of w. If x = and z = , then y is a proper suffix of w.

We extend these denitions to innite words. An innite word w can be written as aninnite sequence of nite subwords ( wn )n 0 = w0w1w2 . A word y is a subword of w if there exist x and z such that w = xyz . If x = , then y is a prex (and a proper prex) of w . A word z is a subword (and a suffix) of w if there exists a y such that w = yz . If y = ,then z is a proper suffix of w .

The set of all nite words made up of letters chosen from an alphabet is denoted by. Note that . If a and w , then |w|a denotes the number of occurrencesof the letter a in the word w. The frequency of a letter a in an innite word w = ( wn )n 0

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(where the ws are letters), denoted by Freq w (a), is limn 1n |w0w1 wn 1 |a , if this limitexists.

Let and be alphabets. A morphism is a map from to that obeys theidentity (wx) = (w)(x) for all words w, x . If = , then the application of amorphism can be iterated. For example, if is the morphism mapping 0 to 1 and 1 to 10 ,then (0 ) = 1 , 2(0 ) = ((0 )) = (1 ) = 10 , and so on. We dene 1(w) = (w) and0(w) = w for any word w.

A morphism : is k-uniform if there is a constant k such that |(a)| = k forall a . A coding is a 1-uniform morphism.

A xed point of a morphism : is a nite word w (or innite word w ) suchthat (w) = w (or (w ) = w ). If there exists a letter a such that (a) = ax and x is aword composed of letters x i such that m (x i ) = for any m Z + , then the morphism is prolongable on the letter a. If so, then lim m m (a) (denoted by (a)) is the xedpoint of iterated on a, where the length of the iterates from the letter a tends to innity.

Let w denote the complement of the nite binary word w (and w denote the complementof the innite binary word w ) where the overbar represents the morphism mapping 0 1and 1 0 . For example, if w = 001 , then w = 110 .

Given a nite word w = a0a1 an , where the as are letters, its reversal , denoted bywR , is an a1a0 .

Denition 1. A generalized choral sequence is an innite binary word c (r 0 , r 1 , r c , z) =(cn )n 0 dened by c3i + r 0 = 0 , c3i + r 1 = 1 , c3i + r c = ci , and c0 = z (where the r s are distinct xed elements of {0, 1, 2} and z = 0 if r 0 = 0 , z = 1 if r 1 = 0 , and z could either be 0 or 1if r c = 0 ) for all i N .

There are eight distinct choral sequences: (Spaces have been inserted to improve read-ability.)

c (0, 2, 1, 0 ) = 001 001 011 001 001 011 001 011 011 c (1, 2, 0, 0 ) = 001 001 101 001 001 101 101 001 101 c (0, 1, 2, 0 ) = 010 011 010 010 011 011 010 011 010 c (2, 1, 0, 0 ) = 010 110 010 110 110 010 010 110 010 c (1, 2, 0, 1 ) = 101 001 101 001 001 101 101 001 101 c (1, 0, 2, 1 ) = 101 100 101 101 100 100 101 100 101 c (2, 1, 0, 1 ) = 110 110 010 110 110 010 010 110 010 c (2, 0, 1, 1 ) = 110 110 100 110 110 100 110 100 100 Sequence c (0, 2, 1, 0 ) is Stewarts choral sequence. (Stewart [11] presented the sequence

(cn )n 1 and not ( cn )n 0 .) Sequence c (0, 1, 2, 0 ) is from Berstel and Karhum aki [2].

Denition 2. A generalized choral sequence c (r 0 , r 1 , r c , z) is called a type-012 sequenceif (r 0 , r 1 , r c ) is a circular permutation of (0, 1, 2). Otherwise (if (r 0 , r 1 , r c ) is a circular permutation of (2, 1, 0)), it is called a type-210 sequence .

2 Some Properties

A previous work [8] presented a characteristic function for generalized choral sequences aswell as proofs of the following two theorems.

Theorem 1. A generalized choral sequence is cube-free , that is, it does not contain any subword of the form xxx , where x is a non-empty nite subword.

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Theorem 2. Given two generalized choral sequences, if they are both type-012 (or if they are both type-210) then any nite subword of one is a subword of the other. Otherwise, any nite subword of one is the complement of a subword of the other.

The following theorem is related to the previous theorem and their proofs are similar.

Theorem 3. Given a type-012 sequence and a type-210 sequence, any nite subword of oneis the reversal of a subword of the other.

Proof. Let v = ( vn )n 0 be the given type-012 sequence and w = ( wn )n 0 be the giventype-210 sequence. We will show that any nite subword of v is the reversal of a subwordof w . (The proof that any nite subword of w is the reversal of a subword of v is similar.)

Because any generalized choral sequence has all the subwords 001 , 010 , 011 , 100 , 101 ,and 110 [8], any length-3 subword of v is the reversal of a subword of w . That is, for agiven j N , there exists a k N such that vj vj +1 vj +2 = wk +2 wk +1 wk .

By Denition 2, there exists a subword v3j + r c v3j + r c +1 v3j + r c +2 v3j + r c +3 v3j + r c +4 v3j + r c +5v3j + r c +6 = vj 01 vj +1 01 vj +2 and a subword w3k + r c w3k + r c +1 w3k + r c +2 w3k + r c +3 w3k + r c +4w3k + r c +5 w3k + r c +6 = wk 10 wk +1 10 wk +2 such that vj 01 vj +1 01 vj +2 = wk +2 01 wk +1 01 wk .Any length-5 subword of v is a subword of a length-7 subword of the form v3j + r c v3j + r c +1v3j + r c +2 v3j + r c +3 v3j + r c +4 v3j + r c +5 v3j + r c +6 . Thus, any length-5 subword of v is the reversalof a subword of w . (This is also true for subwords of length less than 5.)

Extending this reasoning to arbitrarily long nite subwords of similar form yields theresult that any nite subword of v is the reversal of a subword of w .

Remark 1. For any generalized choral sequence c , Freq c (0 ) = 12 and Freq c (1 ) =12 .

We rst show that Freq c (0 ) = 12 . Let c = ( cn )n 0 . Consider the subword c0c1 cn 1with length n. As n , around 13 of the letters of c0c1 cn 1 will be due to the subwordsc3i + r 0 , around 13 will be due to c3i + r 1 , and around

13 will be due to c3i + r c , for i N such

that 3 i + r < n .Since c3i + r 0 = 0 , the number of occurrences of 0 due to the subwords c3i + r 0 approaches

n3 as n . Since c3i + r 1 = 1 , the number of occurrences of 0 due to the subwords c3i + r 0and c3i + r 1 still approaches n3 as n .

Now consider the subwords c3i + r c . Since c3i + r c = ci , one third of the subwords c3i + r cwill be 0 , and one third will be 1 . Of the remaining subwords, one third will be 0 , one thirdwill be 1 and so on.

Thus, the number of occurrences of 0 in the subword c0c1 cn 1 as n approachesn3 1 +

13 1 +

13 (1 + ) =

n3 +

n32 +

n33 + and Freq c (0 ) = lim n

1n

n3 +

n32 +

n33 +

= i =113 i = 1 +

i =1

13

i 1 = 1 + 11 13 =12 .

Using similar reasoning, we can show that Freq c (1 ) = 12 .

Proposition 1. c (1, 2, 0, z) = zc (0, 1, 2, 0 ) and c (2, 1, 0, z) = zc (1, 0, 2, 1 )

Proof. We rst show that c (1, 2, 0, z) = zc (0, 1, 2, 0 ). Let ( an )n 0 = c (1, 2, 0, z), (bn )n 0 =c (0, 1, 2, 0 ), and ( cn )n 0 = zc (0, 1, 2, 0 ). Thus, a3i +1 = 0 , a3i +2 = 1 , a3i = a i , b3i = 0 ,b3i +1 = 1 , and b3i +2 = bi for all i N . Also, ci +1 = bi for all i N so c3i +1 = b3i = 0 ,c3i +2 = b3i +1 = 1 , and c3i +3 = b3i +2 = bi = ci +1 for all i N . Since c3( i +1) = ci +1 for i Nand c30 = c0 , c3i = ci for all i N . Also, a3i +1 = 0 = c3i +1 and a3i +2 = 1 = c3i +2 for alli N .

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We will prove that ( an )n 0 = ( cn )n 0 using induction. Assume an = cn for all n k,where n, k N and k is xed. There are three possibilities: k 0 (mod 3), k 1 (mod3), and k 2 (mod 3). If k 0 (mod 3), then k + 1 = 3 i + 1 for some i N andak +1 = a3i +1 = c3i +1 = ck +1 . If k 1 (mod 3), then k + 1 = 3 i + 2 for some i N andak +1 = a3i +2 = c3i +2 = ck +1 . If k 2 (mod 3), then k + 1 = 3 i for some i Z + andak +1 = a3i = a i = ci = c3i = ck +1 (where a i = ci because i 1 and i 3i 1 = k). Thus,if an = cn for all n k, then an = cn for all n k + 1. Since a0 = c0 = z (that is, an = cnfor all n 0), it follows that an = cn for all n N .

The proof that c (2, 1, 0, z) = zc (1, 0, 2, 1 ) is similar.

Proposition 2. The complement of a generalized choral sequence is also a generalized choral sequence.

Proof. Let c (r 0 , r 1 , r c , z) = ( an )n 0 . Thus, a3i + r 0 = 0 , a3i + r 1 = 1 , and a3i + r c = a i for alli N . Also, a0 = z where z = 0 if r 0 = 0, z = 1 if r 1 = 0, and z = 0 or z = 1 if r c = 0.Let ( bn )n

0= ( an )n

0so that bi = a i for all i N . Thus, b

3i+

r0

= a3

i+

r0

= 1 , b3

i+

r1

=a3i + r 1 = 0 , and b3i + r c = a3i + r c = a i = bi for all i N . Also, b0 = a0 = z where z = 1 if r 0 = 0, z = 0 if r 1 = 0, and z = 1 or z = 0 if r c = 0. Thus, c (r 0 , r 1 , r c , z) = c (r 0 , r 1 , r c , z )where r 0 = r 1 , r 1 = r 0 , r c = r c , and z = 0 if r 0 = 0, z = 1 if r 1 = 0 and z = 0 or z = 1 if r c = 0.

We follow Cassaigne and Karhum aki [3]. Let be an alphabet and ? be a letter notin . For a word w = xy with x and y ( {?}), let T 0(w) = w and let T i (w)for i Z + be the word obtained from T i 1(w) by replacing the rst occurrence of ? inT i 1(w) by the i-th letter of T i 1(w). The Toeplitz word determined by the pattern w isT (w) = lim i T i (w), an innite word over .

Using this denition of Toeplitz words and Denition 1, we see that some generalizedchoral sequences are Toeplitz words over {0 ,1 } [3, Example 4].

Remark 2. The sequences c (0, 2, 1, 0 ), c (0, 1, 2, 0 ), c (1, 0, 2, 1 ), and c (2, 0, 1, 1 ) are theToeplitz words T (0 ?1 ), T (01 ?), T (10 ?), and T (1 ?0 ), respectively.

The other generalized choral sequences are not Toeplitz words (unless we relax the con-dition that w start with a letter from [3, Example 2]).

2.1 Automatic Sequence

We follow Allouche and Shallit [1] again. A deterministic nite automaton with output (DFAO) is a model of computation dened by a nite set of states Q, a nite input alphabet, a transition function : Q Q (which we extend to : Q Q so that (q, ) = qand (q,xa ) = ((q, x), a) for all q Q, x , and a [1, p. 129]), an initial stateq0 Q, a nite output alphabet , and an output function : Q . A DFAO has a wordw as input and it moves from state to state according to while reading the lettersof w (in order from left to right). When the end of w is reached, the automaton halts in astate q and outputs the letter (q).

A DFAO with representations of base- k numbers as input is called a k-DFAO. A nite-state function f : is one that can be computed by a DFAO such that f (w) = ((q0 , w)). Informally, a word w = ( wn )n 0 is k-automatic if wn is a nite-state functionof the base- k digits of n (starting with the most signicant digit). (Note that the inputcan have an arbitrary nite number of leading zeros.) A k-automatic innite word has anassociated k-DFAO.

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Proposition 4. A generalized choral sequence c (r 0 , r 1 , r c , z) = ( cn )n 0 is the xed point of the morphism iterated on z mapping 0 a0a1a2 and 1 b0b1b2 , where a r 0 = br 0 = 0 ,a r 1 = br 1 = 1 , a r c = 0 , and br c = 1 .

Proof. Recall that z = 0 if r 0 = 0, z = 1 if r 1 = 0, and z could either be 0 or 1 if r c = 0. If

z = 0 , then either r 0 = 0 or r c = 0. Either way, a0 = 0 and 0 maps to 0 a1a2 . If z = 1 , theneither r 1 = 0 or r c = 0. Either way, b0 = 1 and 1 maps to 1 b1b2 . Since a1 , a 2 , b1 , b2 {0 , 1 },the morphism is prolongable on the letter z. Thus, iterated on z has a unique xedpoint (z) which we will call w = ( wn )n 0 , where the ws are letters. Note that w0 = z.

The morphism is 3-uniform (because (0 ) = a0a1a2 and (1 ) = b0b1b2). Thus (by alemma in [1, p. 174]), (wi ) = w3i +0 w3i +1 w3i +2 for all i N . If r 0 = 0, then (0 ) = 0 a1a2and (1 ) = 0 b1b2 . If r 0 = 1, then (0 ) = a0 0 a2 and (1 ) = b0 0 b2 . If r 0 = 2, then(0 ) = a0a1 0 and (1 ) = b0b1 0 . In any case, w3i + r 0 = 0 for all i N . Similarly, it can beseen that w3i + r 1 = 1 for all i N . If r c = 0, then (0 ) = 0 a1a2 and (1 ) = 1 b1b2 . If r c = 1,then (0 ) = a0 0 a2 and (1 ) = b0 1 b2 . If r c = 2, then (0 ) = a0a1 0 and (1 ) = b0b1 1 . Inany case, w3i + r c = wi for all i N . By Denition 1, c = w = (z).

A morphism : is said to be primitive if there exists an integer n 1 such thatfor all a, b , a occurs in n (b). The morphism in Proposition 4 is primitive because 0occurs in n (1 ) and 1 occurs in n (0 ) for any n Z + .

An innite word w is uniformly recurrent if, for every nite subword y of w , there existsan integer k such that every subword of length k of w contains y.

If is a primitive morphism prolongable on z, then (z) is uniformly recurrent [1,Theorem 10.9.5].

Remark 3. A generalized choral sequence is uniformly recurrent.

2.3 Subword Complexity

Let Sub w (n) denote the set of all subwords of length n of an innite word w and pw (n)denote the subword complexity function of w , the function counting the number of distinctlength- n subwords of w .

For a generalized choral sequence c , Sub c (1) = {0 , 1 }, Sub c (2) = {00 , 01 , 10 , 11 }, andSub c (3) = {001 , 010 , 011 , 100 , 101 , 110 }. Thus, pc (1) = 2, pc (2) = 4, and pc (3) = 6. If cis a type-012 sequence, then Sub c (4) = {0010 , 0011 , 0100 , 0110 , 1001 , 1010 , 1011 , 1101 }; if

c is a type-210 sequence, then Sub c (4) = {0010 , 0100 , 0101 , 0110 , 1001 , 1011 , 1100 , 1101 }.In either case, pc (4) = 8.

Theorem 4. For a generalized choral sequence c , pc (n) = 2 n for all n Z + .

Proof. Let S n = {cj cj +1 cj + n 1 : j N } be the set of all length- n subwords of a gener-alized choral sequence c (r 0 , r 1 , r c , z) = ( cn )n 0 . The set S n can be partitioned into threesubsets: S n, 0 contains only all the subwords with initial index j r 0 (mod 3), S n, 1 containsthose with j r 1 (mod 3), and S n,c contains those with j r c (mod 3). Let |S | denote thenumber of elements in a set S . Thus, pc (n) = |S n | = |S n, 0 | + |S n, 1 | + |S n,c |.

The proof is by induction: given pc (i) = 2 i for all i k where i, k Z + and k is xed,we will show that pc (i) = 2 i for all i k + 1. It is easy to see that pc (i) = 2 i for all i 3,so we only need to look at k 3.

If c is, say, a type-012 sequence then length- k subwords of c can be visualized as shownbelow, where the bottom overline shows a subword from S k, 0 , the middle overline shows

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a subword from S k, 1 , and the top overline shows a subword from S k,c . (If c is a type-210sequence, then the subwords would be of the form 10 ci 10 ; the end result is the same.)

If k 0 (mod 3): 0 1 ci 0 1 ci + m 1 0 1 ci + m 0 1

If k 1 (mod 3): 0 1 ci 0 1 ci + m 1 0 1 ci + m 0 1 If k 2 (mod 3): 0 1 ci 0 1 ci + m 1 0 1 ci + m 0 1

Note that 0 < 3m k since k 3. Thus, m < k and m + 1 < k .We are given |S k, 0 | + |S k, 1 | + |S k,c | = 2 k. Consider what happens when k 0 (mod 3)

and we now look at subwords of length k + 1:

0 1 ci 0 1 ci + m 1 0 1 ci + m 0 1

All the subwords in S k +1 ,0 are just the subwords in S k, 0 with the letter 0 concatenated

at the end; the number of subwords remains the same. That is, |S k +1 ,0 | = |S k, 0 | . Allthe subwords in S k +1 ,1 are just those in S k, 1 with a 1 concatenated at the end. Thus,|S k +1 ,1 | = |S k, 1 | .

Note that |S k,c | is equal to the number of distinct subwords ci ci +1 ci + m 1 , that is,|S k,c | = |S m,c |. Since m < k , |S m,c | = 2 m. Also, |S k +1 ,c | is equal to the number of distinctsubwords ci ci +1 ci + m , that is, |S k +1 ,c | = |S m +1 ,c | = 2( m + 1) = 2 m + 2 = |S m,c | + 2 =|S k,c | + 2. (Since m + 1 < k , |S m +1 ,c | = 2( m + 1).) Thus, |S k +1 | = |S k +1 ,0 | + |S k +1 ,1 | +|S k +1 ,c | = |S k, 0 |+ |S k, 1 |+( |S k,c |+2) = |S k |+2 = 2 k+2 = 2( k+1), that is, pc (k+1) = 2( k+1).Thus, pc (i) = 2 i for all i k + 1.

Using similar reasoning, we nd that when k 1 (mod 3), |S k +1 ,0 | + |S k +1 ,1 | + |S k +1 ,c | =

|S k, 0 | + ( |S k, 1 | + 2) + |S k,c |; when k 2 (mod 3), |S k +1 ,0 | + |S k +1 ,1 | + |S k +1 ,c | = ( |S k, 0 | +2) + |S k, 1 | + |S k,c | . In any case, the end result is the same.

It is easy to see that pc (i) = 2 i for all i 3, i Z + . Thus, pc (n) = 2 n for all n Z + .

2.4 Lyndon Factorization

We now follow Richomme [9]. Words may be ordered lexicographically. Let the alphabet{0,1 } be ordered such that 0 < 1 . We say that v w (or w v) if and only if either v isa prex of w or there exist words x,y,z and letters a, b such that v = xay , w = xbz, anda < b . We say that v < w (or w > v ) if v w and v = w.

Remark 4. [4, p. 82] Let a, b, c, and d be nite words over an ordered alphabet. If a < band |a | | b| , then ac < bd .

We extend lexicographic order to the set of nite or innite words [6]. We say that v < wif and only if either v is a prex of w or there exist words x,y, z and letters a, b such thatv = xay , w = xbz , and a < b . We say that v < w if and only if there exist words x, y , zand letters a, b such that v = xa y , w = xbz, and a < b . We say that v < w if and only if there exist words x, y , z and letters a, b such that v = xa y , w = xbz , and a < b .

Chen, Fox, and Lyndon [4] introduced what they called standard sequences but which arenow called Lyndon words . A Lyndon word is a word that is less than any of its non-emptyproper suffixes. Lyndon words were originally dened as nite words [4] but the denitionwas eventually extended to include innite words [10, Proposition 2.2]. For example, lettersare Lyndon words; 01011 is a nite Lyndon word while 01101 is not; and 01 = 0111 isan innite Lyndon word while ( 01 ) = 010101 is not.

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Lemma 1. [10] An innite word is an innite Lyndon word if and only if it has an innitenumber of prexes which are Lyndon words.

A morphism over an ordered alphabet is order-preserving if for all u, v , u vimplies (u) (v) [9]. For an order-preserving morphism , if u < v then (u) < (v) [9,

Lemma 3.2].Remark 5. For an order-preserving morphism , if u < v then n (u) < n (v) for n Z + .

A morphism is a Lyndon morphism [9] if it preserves (nite) Lyndon words. A morphism on {0 ,1 } such that 0 < 1 is a Lyndon morphism if and only if (0 ) and (1 ) are Lyndonwords and (0 ) < (1 ) [9, Proposition 4.7].

Remark 6. The morphism mapping 0 001 and 1 011 is a Lyndon morphism.

Lemma 2. For the morphism mapping 0 001 and 1 101 , n (0 i 1 j ) for n, j N ,

i Z +

is a Lyndon word.Proof. We rst show that n (0 ) for n N is a Lyndon word (the case where i = 1 and j = 0).

The words (0 ) = 001 and (1 ) = 101 differ only in the rst letter. The words 2(0 ) =(0 )(0 )(1 ) and 2 (1 ) = (1 )(0 )(1 ) also differ only in the rst letter because (0 ) and(1 ) differ only in the rst letter. Continuing this reasoning, it can be seen that n (0 ) andn (1 ) for n Z + differ only in the rst letter and that if n (0 ) = 0 x then n (1 ) = 1 x.

Assume that k (0 ) is a Lyndon word for some k Z + . Now, k +1 (0 ) = k (0 )k (0 )k (1 )= 0 x0 x1 x for some x = a1a2 an where the as are letters. Since k (0 ) is a Lyndon word,it is less than any of its non-empty proper suffixes. Thus, 0 x < a 1a2 an , 0 x < a 2 an ,and so on up to 0 x < a n .

Using Remark 4 and starting from 0 x < x , we get 0 x0 x1 x < x 0 x1 x. From 0 x < a 2 an ,we get 0 x0 x1 x < a 2 an 0 x1 x. Similarly, we go on up to 0 x0 x1 x < a n 0 x1 x.

Clearly, 0 x0 x1 x < 0 x1 x. Using Remark 4 and starting from 0 x < x , we get 0 x0 x1 x n +1 (1 ) for n N .

Proof. By Remark 6, is a Lyndon morphism. Thus, it is order-preserving [9, Proposition4.2]. From Remark 5, since 011 < 1 , then n (011 ) = n +1 (1 ) < n (1 ) for n N .

Lemma 4. For the morphism mapping 0 001 and 1 101 , n (01 m ) > n +1 (01 m )

for n, m Z+

.Proof. The morphism is order-preserving because (01 ) < (1 ) [9, Lemma 3.13]. Fixm Z + . From Remark 5, since 001 (101 )m < 01 m , then n (001 (101 )m ) = n +1 (01 m ) 1 =001101 . From Lemma 4, 1 > 2 > 3 > .

The proof of Proposition 7 is relatively detailed. For brevity, we omit some details inthe remaining proofs.

Proposition 8. The sequence c (2, 1, 0, 0 ) is an innite non-increasing sequence of niteLyndon words ( k )k 0 with 0 = 01011 and k = k (011 ) for k Z + , where is themorphism mapping 0 001 and 1 011 .

Proof. Let c (2, 1, 0, 0 ) = ( cn )n 0 and ( k )k 0 = ( wn )n 0 where cn , wn {0 , 1 }. Now,

c3i = ci , c3i +1 =1

, c3i +2 =0

, and c0 =0

for i N

. The wn s are characterized byw0w1w2w3w4 = 01011 , wm +3 p = 0 , wm +3 p+1 = wm + p , and wm +3 p+2 = 1 where m =2 + kj =1 3

j and m = 2 + k 1j =1 3j , for all p N such that p 3k 1, and for all k Z + .

Clearly, c0c1c2c3c4 = w0w1w2w3w4 . Because m 2 (mod 3), it follows that c3i = wm +3 p+1 ,c3i +1 = wm +3 p+2 , and c3i +2 = wm +3 p .

Subword 0 is a Lyndon word and so is each k for k Z + because is a Lyndonmorphism (Remark 6) and 011 is a Lyndon word. Clearly, 0 = 01011 > 1 = 001011011 .Note that k = k +1 (1 ). From Lemma 3, 1 > 2 > 3 > .

Proposition 9. The sequence c (1, 2, 0, 1 ) is an innite non-increasing sequence of nite

Lyndon words ( k )k 0 with 0 = 1 , 1 = 01 , and k = k 1

( 1) for k 2, where is themorphism mapping 0 001 and 1 101 .

Proof. Let c (1, 2, 0, 1 ) = ( cn )n 0 and ( k )k 0 = ( wn )n 0 where cn , wn {0 , 1 }. Now,c3i = ci , c3i +1 = 0 , c3i +2 = 1 , and c0 = 1 for i N . The wn s are characterized by

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w0w1w2w3 = 1010 , wm +3 p = wm + p , wm +3 p+1 = 0 , and wm +3 p+2 = 1 where m = 1 +kj =2 2 3

j 2 and m = 1 + k 1j =2 2 3j 2 , for all p N such that p 2 3k 2 1, and for

all k 2. Clearly, c0c1c2 = w0w1w2 . Because m = 3 +kj =3 2 3

j 2 and m 0 (mod 3),it follows that c3i = wm +3 p , c3i +1 = wm +3 p+1 , and c3i +2 = wm +3 p+2 .

Subword 0 is a Lyndon word and, by Lemma 2, so is each k +1 = k

(01 ) for k N

.Clearly, 0 = 1 > 1 = 01 > 2 = 001101 . From Lemma 4, 2 > 3 > 4 > .

Proposition 10. The sequence c (1, 0, 2, 1 ) is an innite non-increasing sequence of niteLyndon words ( k )k 0 with 0 = 1 and k = k ( 0) for k Z + , where is the morphism mapping 0 001 and 1 011 .

Proof. Let c (1, 0, 2, 1 ) = ( cn )n 0 and ( k )k 0 = ( wn )n 0 where cn , wn {0 , 1 }. Now, c3i =1 , c3i +1 = 0 , and c3i +2 = ci for i N . The wn s are characterized by w0 = 1 , wm +3 p = 0 ,wm +3 p+1 = wm + p , and wm +3 p+2 = 1 where m =

kj =1 3

j 1 and m = k 1j =1 3j 1 , for

all p N such that p 3k 1 1, and for all k Z + . Clearly, c0 = w0 . Becausem = 1 + kj =2 3

j 1 and m 1 (mod 3), it follows that c3i = wm +3 p+2 , c3i +1 = wm +3 p , andc3i +2 = wm +3 p+1 .

Subword 0 is a Lyndon word and so is each k for k Z + because is a Lyndonmorphism (Remark 6). From Lemma 3, 0 > 1 > 2 > .

Proposition 11. The sequence c (2, 1, 0, 1 ) is an innite non-increasing sequence of niteLyndon words ( k )k 0 with 0 = 1 = 1 and k = k 1( 1) for k 2, where is themorphism mapping 0 001 and 1 011 .

Proof. Let c (2, 1, 0, 1 ) = ( cn )n 0 and ( k )k 0 = ( wn )n 0 where cn , wn {0 , 1 }. Now,c3i = ci , c3i +1 = 1 , c3i +2 = 0 , and c0 = 1 for i N . The wn s are characterized byw0w1 = 11 , wm +3 p = 0 , wm +3 p+1 = wm + p , and wm +3 p+2 = 1 where m = 1 +

kj =2 3

j 2

and m = 1 + k 1j =2 3j 2 , for all p N such that p 3k 2 1, and for all k 2. Clearly,

c0c1 = w0w1 . Because m = 2+kj =3 3

j 2 and m 2 (mod 3), it follows that c3i = wm +3 p+1 ,c3i +1 = wm +3 p+2 , and c3i +2 = wm +3 p .

Subwords 0 and 1 are Lyndon words and so is each k for k 2 because is a Lyndonmorphism (Remark 6). Note that 0 = 1 = 1 . From Lemma 3, 1 > 2 > 3 > .

Proposition 12. The sequence c (2, 0, 1, 1 ) is an innite non-increasing sequence of niteLyndon words ( k )k 0 with 0 = 1 = 1 , 2 = 011 , 3 = 01 , and k = ( k 2) for k 4,where is the morphism mapping 0 001 and 1 101 .

Proof. Let c (2, 0, 1, 1 ) = ( cn )n 0 and ( k )k 0 = ( wn )n 0 where cn , wn {0 , 1 }. Now,c3i = 1 , c3i +1 = ci , and c3i +2 = 0 for i N .

Note that for k = 2 j + 2 and j Z + , k is an even number greater than or equal to 4.For k = 2 j + 3 and j Z + , k is an odd number greater than or equal to 5.

From 2j +2 = j (011 ) for j Z + , we get | 2j +2 | = 3 j +1 . Thus, 2j +2 = wq wq+1 wq+3 j +1 1 where q = 2 +

ji =1 (3

i + 2 3i 1). Also, wq+3 p = wq + p , wq+3 p+1 = 0 , andwq+3 p+2 = 1 where q = 2 + j 1i =1 (3 i + 2 3i 1) and p goes from 0 to 3j 1.

From 2j +3 = j (01 ) for j Z + , we get | 2j +3 | = 2 3j . Thus, 2j +3 = wr wr +1 wr +2 3 j 1 where r = 5 +

ji =1 (3

i +1 + 2 3i 1). Also, wr +3 p = wr + p , wr +3 p+1 = 0 , andwr +3 p+2 = 1 where r = 5 +

j 1i =1 (3

i +1 + 2 3i 1) and p goes from 0 to 2 3j 1 1.

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Now, c0c1c2c3c4c5c6 = w0w1w2w3w4w5w6 . Note that q = 7 +ji =2 (3

i + 2 3i 1 ) andr = 16 + ji =2 (3

i +1 + 2 3i 1) so that q r 1 (mod 3). Thus, it follows that c3i =wq+3 p+2 = wr +3 p+2 , c3i +1 = wq+3 p = wr +3 p , and c3i +2 = wq+3 p+1 = wr +3 p+1 .

Subwords 0 and 1 are Lyndon words and, by Lemma 2, so is each 2j +2 = j (011 ) andeach 2j +3 = j (01 ) for j N . Clearly, 0 = 1 = 1 > 2 = 011 . Note that j (01 )j (1 ) =j (011 ) > j (01 ) for j N . Thus, 2j +2 > 2j +3 for j N , that is, 2 > 3 , 4 > 5 ,and so on. The morphism is order-preserving because (01 ) < (1 ) [9, Lemma 3.13].From Remark 5, since 01 > 001101101 , then j (01 ) > j (001101101 ) = j +1 (011 ) for j N . Thus, 2j +3 > 2j +4 for j N , that is, 3 > 4 , 5 > 6 , and so on. We then get

2 > 3 > 4 > 5 > 6 > .

Acknowledgment

I thank an anonymous colleague for pointing out the connection to Toeplitz words.

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[6] Guy Melancon. Lyndon factorization of innite words. In STACS 96 , volume 1046/1996of Lecture Notes in Computer Science , pages 147154. Springer Berlin/Heidelberg,1996.

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on generalized choral sequences

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